Chapter 4.4, Problem 17E

To determine

To find: The ratio of $h$ to $r$ .

Answer to Problem 17E

  $\frac{8}{\pi }\text{ to 1}$

Explanation of Solution

Given information: The volume of open circular cylindrical can is $1000{\text{ cm}}^3$ .

Concept used: The maximum or minimum value of the function $f\left(x\right)$ will be at the critical point of that function.

Calculation:

Consider that the radius of the base of the tank is $r$ and its height is $h$ .

The volume of the open tank (V) is written as shown below.

  $V=\pi \left(r^2\right)\left(h\right)$

Substitute $V=1000$ into the obtained equation and solve for the expression of $h$ .

  $\begin{array}{c}1000=\pi \left(r^2\right)\left(h\right)\\ h=\frac{1000}{\pi r^2}\end{array}$

Now, write the total surface Area (S) for the given statement.

  $\begin{array}{c}S=8r^2+2\pi rh\\ =8r^2+2\pi r\left(\frac{1000}{\pi r^2}\right)\\ =8r^2+\left(\frac{2000}{r}\right)\end{array}$

Differentiate the obtained equation with respect to $r$ .

  $\frac{dS}{dr}=16r-\left(\frac{2000}{r^2}\right)$

Equate the obtained derivative to 0 and solve for $r$ .

  $\begin{array}{c}\frac{dS}{dr}=0\\ 16r-\left(\frac{2000}{r^2}\right)=0\\ r=5\end{array}$

Now, differentiate the equation, $\frac{dS}{dr}=16r-\left(\frac{2000}{r^2}\right)$ with respect to $r$ .

  $\frac{d^{2}S}{d{r}^2}=16+\frac{4000}{r^3}$

Since at $r=5$ , $\frac{d^{2}S}{d{r}^2}>0$ , the volume will be minimum at $r=5$ .

Substitute $r=5$ into the equation, $h=\frac{1000}{\pi r^2}$ , and solve for $h$ .

  $\begin{array}{c}h=\frac{1000}{\pi {\left(5\right)}^2}\\ =40{\left(\pi \right)}^{-1}\end{array}$

Find the ratio of $h$ to $r$ .

  $\begin{array}{c}\frac{h}{r}=\frac{40{\left(\pi \right)}^{-1}}{5}\\ =\frac{8}{\pi }\end{array}$

So, the required ratio is $\frac{8}{\pi }\text{ to 1}$.